Question

The resistance of a wire of length \( L \) and radius \( r \) is \( R \). Which one of the following would provide a wire of the same material with resistance \( \frac{R}{2} \)?

• A. Using a wire of same radius and twice the length
• B. Using a wire of same radius and half length
• C. Using a wire of same length and twice the radius
• D. Using a wire of same length and half the radius

Hint:

Remember, the resistance of a wire is inversely proportional to the square of its radius. So, reducing the radius reduces the resistance.

Answer 1

The Correct Option is D

The resistance \( R \) of a wire is given by the formula: 

\( R = \rho \frac{L}{A} \)

where \(\rho\) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

The cross-sectional area \( A \) of a wire with radius \( r \) is:

\( A = \pi r^2 \)

So, the resistance can be expressed as:

\( R = \rho \frac{L}{\pi r^2} \)

To achieve \( \frac{R}{2} \), the new resistance must be half of the original; hence:

\( \frac{R}{2} = \rho \frac{L'}{\pi (r')^2} \)

Let us analyze the options:

• Using a wire of same radius and twice the length: \( R' = \rho \frac{2L}{\pi r^2} = 2R \)
• Using a wire of same radius and half length: \( R' = \rho \frac{L/2}{\pi r^2} = \frac{R}{2} \)
• Using a wire of same length and twice the radius: \( R' = \rho \frac{L}{\pi (2r)^2} = \frac{R}{4} \)
• Using a wire of same length and half the radius: \( R' = \rho \frac{L}{\pi (r/2)^2} = 4R \)

Therefore, the correct choice is: Using a wire of same radius and half length.